Collatz Conjecture

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Collatz Conjecture

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Asset Durmagambetov^*

,Aniyar Durmagambetova

Asset Durmagambetov^*

,Aniyar Durmagambetova

This version is not peer-reviewed

Submitted:

08 January 2024

Posted:

08 January 2024

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Abstract

This paper presents an analysis of the number of zeros in the binary representation of natural numbers. The primary method of analysis involves the use of the concept of the fractional part of a number, which naturally emerges in the determination of binary representation. This idea is grounded in the fundamental property of the Riemann zeta function, constructed using the fractional part of a number. Understanding that the ratio between the fractional and integer parts of a number, analogous to the Riemann zeta function, reflects the profound laws of numbers becomes the key insight of this work. The findings suggest a new perspective on the interrelation between elementary properties of numbers and more complex mathematical concepts, potentially opening new directions in number theory and analysis."

Keywords:

Subject: Computer Science and Mathematics - Analysis

1. Introduction

We will use the following well-known fact: if, for the members of the Collatz sequence, zeros predominate in their binary representation, then these members will lead to a decrease in the subsequent members according to the Collatz rule. A striking example is when the initial number in the Collatz sequence is equal to

2^{n}

. Let’s write the solution of the equation

n = 2^{x}

in the form

x = x + [x]

and note that the smaller x, the more zeros in the corresponding binary representation for n. Developing this idea, we come to the following steps.

Analysis of the binary representation of simple cases of natural numbers.
Creation of a process for decomposing an arbitrary natural number into powers of two.
Analysis of the proximity of the process to binary decomposition at the completion of decomposition at each stage.
Calculation of the number of zeros in the binary decomposition of a natural number.
Estimation of the Collatz sequence members depending on the number of ones in the binary decomposition.

2. Results

This document reveals a comprehensive solution to the Collatz Conjecture, as first proposed in [1]. The Collatz Conjecture, a well-known unsolved problem in mathematics, questions whether iterative application of two basic arithmetic operations can invariably convert any positive integer into 1. It deals with integer sequences generated by the following rule: if a term is even, the subsequent term is half of it; if odd, the next term is the previous term tripled plus one. The conjecture posits that all such sequences culminate in 1, regardless of the initial positive integer.

Named after mathematician Lothar Collatz, who introduced the concept in 1937, this conjecture is also known as the 3n + 1 problem, the Ulam conjecture, Kakutani’s problem, the Thwaites conjecture, Hasse’s algorithm, or the Syracuse problem. The sequence is often termed the hailstone sequence due to its fluctuating nature, resembling the movement of hailstones.

Paul Erdős and Jeffrey Lagarias have commented on the complexity and mathematical depth of the Collatz Conjecture, highlighting its challenging nature.

Consider an operation applied to any positive integer:

Divide it by two if it’s even.
Triple it and add one if it’s odd.

This operation is mathematically defined as:

f (n) = \{\begin{matrix} \frac{n}{2}, & if n \equiv 0 mod 2, \\ 3 n + 1, & if n \equiv 1 mod 2 . \end{matrix}

A sequence is formed by continuously applying this operation, starting with any positive integer, where each step’s result becomes the next input. The Collatz Conjecture asserts that this sequence will always reach 1 Recent substantial advancements in addressing the Collatz problem have been documented in works [2].

Now let’s move on to our research, which we will conduct according to the announced plan. For this, we will start with the following

Theorem 1.

Let

x \in N, [α_{j}] - [α_{j + 1}] = δ_{j} > 0, ϵ_{1} < 1 / 2,

α_{j} = [α_{j}] + ϵ_{j}, ϵ_{j} < 1,

x = \sum_{i = 1}^{j - 1} 2^{[α_{i}]} + 2^{α_{j}}, x = \sum_{i = 1}^{j} 2^{[α_{i}]} + 2^{α_{j + 1}}, σ_{j} = 1 - ϵ_{j}

Then

δ_{j} = 1

σ_{j + 1} l n 2 = \frac{2 σ_{j} l n 2}{1 - σ_{j + 1} l n 2} + o (σ_{j + 1}^{2} / 4)

δ_{j} > 1

σ_{j + 1} l n 2 = - 2^{δ_{j} - 1} \frac{l n 2 - 2^{- δ_{j} - 1}}{1 - σ_{j + 1} l n 2 / 2} + 2^{δ_{j} - 1} σ_{j} l n 2 \frac{1}{1 - σ_{j + 1} l n 2 / 2} + o (σ_{j + 1}^{2})

Proof.

2^{ϵ_{j}} = 2^{- δ_{j} + ϵ_{j + 1}} + 1 \Rightarrow 2^{1 - σ_{j}} = 2^{- δ_{j} + 1 - σ_{j + 1}} + 1 \Rightarrow

l n (2^{1 - σ_{j}}) = l n 2 - σ_{j} l n 2 = l n (2^{- δ_{j} + 1 - σ_{j + 1}} + 1)

Computing as

δ_{j} = 1

l n (2^{- δ_{j} + 1 - σ_{j + 1}} + 1) |_{δ_{j} = 1} = l n (2^{- σ_{j + 1}} + 1) = l n ((1 - σ_{j + 1} l n 2 + σ_{j + 1}^{2} l n^{2} / 2) + 1 + o (σ_{j + 1}^{2} / 4))

l n (2 - σ_{j + 1} l n 2 + σ_{j + 1}^{2} l n 2^{2} / 2) = l n 2 + l n (1 - σ_{j + 1} l n 2 / 2 + σ_{j + 1}^{2} l n^{2} / 4 + o (σ_{j + 1}^{2} / 4))

l n (2^{- σ_{j + 1}} + 1) = l n 2 - l n 2 σ_{j + 1} / 2 + l n^{2} 2 σ_{j + 1}^{2} / 2 + o (σ_{j + 1}^{2} / 4)

l n 2 - σ_{j} l n 2 = l n 2 - l n 2 σ_{j + 1} / 2 + l n^{2} 2 σ_{j + 1}^{2} / 4 + o (σ_{j + 1}^{2} / 4)

σ_{j + 1} l n 2 = \frac{2 σ_{j} l n 2}{1 - σ_{j + 1} l n 2 / 2} + o (σ_{j + 1}^{2} / 4)

Repeating computing as

δ_{j} > 1

we get

l n (2^{- δ_{j} + 1 - σ_{j + 1}} + 1) = l n (2^{- δ_{j} + 1} 2^{- σ_{j + 1}} + 1) =

l n (1 + 2^{- δ_{j} + 1} + 2^{- δ_{j} + 1} [- σ_{j + 1} l n 2 + σ_{j + 1}^{2} l n 2^{2} / 2] + o (σ_{j + 1}^{2} / 4 + 2^{- δ_{j} + 1}) =

2^{- δ_{j} + 1} + 2^{- δ_{j} + 1} [- σ_{j + 1} l n 2 + σ_{j + 1}^{2} l n 2^{2} / 2] + o (σ_{j + 1}^{2} + 2^{- δ_{j} + 1}) \Rightarrow

l n 2 - σ_{j} l n 2 = 2^{- δ_{j} + 1} + 2^{- δ_{j} + 1} [- σ_{j + 1} l n 2 + σ_{j + 1}^{2} l n 2^{2} / 2] + o (σ_{j + 1}^{2} + 2^{- δ_{j} + 1})

σ_{j + 1} l n 2 = - 2^{δ_{j} - 1} \frac{l n 2 - 2^{- δ_{j} + 1}}{1 - σ_{j + 1} l n 2 / 2} + 2^{δ_{j} - 1} σ_{j} l n 2 \frac{1}{1 - σ_{j + 1} l n 2 / 2} + o (σ_{j + 1}^{2} + 2^{- δ_{j} + 1})

□

Theorem 2.

Let

x \in N, α_{j} = [α_{j}] x = \sum_{i = 1}^{j - 1} 2^{[α_{i}]} + 2^{α_{j}}

Then the number of zeros in the binary representation

C_{z}

is calculated by the following formula

C_{z} = \sum_{i = 1}^{j - 1} [δ_{i} - 1] + α_{j} - 1

Proof.

C_{z} = \sum_{i = 1}^{j - 1} [α_{i} - α_{i + 1} - 1] + α_{j} - 1

By definition

δ_{i}

C_{z} = \sum_{i = 1}^{j - 1} [δ_{i} - 1] + α_{j} - 1

□

Let’s introduce

μ_{k}, ν_{k}

for

x = \sum_{i = 0}^{n} γ_{i} 2^{i}

by following rules

γ_{k} + γ_{k + 1} = 1, γ_{k + μ_{k}} + γ_{k + μ_{k} + 1} = 1, \prod_{i = k + 1}^{i = μ_{k}} γ_{i} = 1;

γ_{j} + γ_{j + 1} = 1, γ_{j + μ_{j}} + γ_{j + ν_{j} + 1} = 1, ν_{j} = \sum_{i = j + 1}^{i = ν_{j}} (1 - γ_{i})

another words

μ_{k},

is count of ones starting at point k with no zeros in between until the first zero or until the end of the sequence

ν_{j}

is count of zeros starting at point j with no ones in between until the first zero or until the end of the sequence

Theorem 3.

Let

x = 3^{n} = 2^{[α] + {α}} = \sum_{i = 1}^{n *} γ_{i} 2^{i},

{α} > l n 2, n^{*} = n * [ln (3) / ln (2)]

(1)

then

\sum_{γ_{i} = 0} 1 \geq n^{*} / 2 - 5

Proof.

3^{n} = 2^{α} \Rightarrow α = n / ln (3) / ln (2) \Rightarrow 3^{n} = 2^{[α] + {α}}

Using Theorem 1, we create a sequence

ϵ_{i}, m_{i}, ϵ_{1} = {α}

2^{ϵ_{1}} = \sum_{k = 0}^{i - 1} 2^{[α_{k}] - α_{1}} + 2^{α_{i} - α_{1}}

Suppose

\sum_{γ_{i} = 0} 1 = 0

then by Theorem 1

\Rightarrow σ_{j + 1} l n 2 = \frac{2 σ_{j} l n 2}{1 - σ_{j + 1} l n 2} + o (ln 2 σ_{j + 1}^{2} / 4) \Rightarrow

2^{- 1} σ_{j + 1} l n 2 = \frac{σ_{j} l n 2}{1 - σ_{j + 1} l n 2} + 2^{- 1} * o (ln 2 σ_{j + 1}^{2} / 4)

After repeating j times we get

2^{- j} σ_{j + 1} l n 2 = \frac{σ_{1} l n 2}{\prod_{1}^{j} (1 - σ_{k + 1} l n 2 / 2)} + \sum_{1}^{j} 2^{- k} * o (ln 2 σ_{k + 1}^{2} / 4)

By Theorems (1-2) and condition of the current Theorem proceed

l n 2 / 2 < σ_{1} l n 2 < o (ln 2 σ_{k + 1}^{2} / 4)

immediately

\Rightarrow \sum_{γ_{i} = 0} 1 > 0

Let’s introduce

a s δ_{k} = 1 : α_{k} = 0, β_{k} = \frac{1}{1 - σ_{j + 1} l n 2}

a s δ_{k} > 1 : α_{k} = - 2^{δ_{j} - 1} \frac{l n 2 - 2^{- δ_{j} - 1}}{1 - σ_{j + 1} l n 2 / 2}, β_{k} = \frac{2^{δ_{j} - 1}}{1 - σ_{j + 1} l n 2 / 2}

l n 2 σ_{k + 1} = α_{k} + β_{k} l n 2 σ_{k}

σ_{j + 1} l n 2 = α_{j} + \sum_{m = 1}^{m = j - 1} α_{j - m} \prod_{l = 1}^{l = m} β_{j - l + 1} + \prod_{l = 0}^{l = j - 1} β_{j - l} l n 2 σ_{1} \Rightarrow

\frac{σ_{j + 1} l n 2}{\prod_{l = 0}^{l = j - 1} β_{j - l}} = \sum_{m = 0}^{m = j - 1} \frac{α_{j - m}}{\prod_{l = m + 1}^{l = j - 1} β_{j - l}} \geq σ_{1} l n 2 \Rightarrow

By condition the theorem

\frac{σ_{j + 1} l n 2}{\prod_{l = 0}^{l = n - 1} β_{n - l}} - \sum_{m = 0}^{m = n - 1} \frac{α_{j - m}}{\prod_{l = m}^{l = n - 1} β_{j - l}} \geq σ_{1} l n 2 \Rightarrow

\frac{σ_{n} l n 2}{\prod_{l = 0}^{l = n - 1} β_{n - l}} - \sum_{m = 0}^{m = n - 1} \frac{l n 2 - 2^{- δ_{j} + 1}}{\prod_{l = m + 1}^{l = n - 1} β_{j - l}} \geq σ_{1} l n 2 \Rightarrow

Suppose

\forall i \in (1, n) δ_{i} = 2 \Rightarrow

\frac{σ_{n} l n 2}{\prod_{l = 0}^{l = n - 1} β_{n - l}} + \sum_{m = 0}^{m = n - 1} \frac{l n 2 - 1 / 2}{2^{m}} > σ_{1} l n 2 \Rightarrow

2 (l n 2 - 1 / 2) > σ_{1} \Rightarrow

\exists δ_{j} > 2

In common case

\frac{σ_{n} l n 2}{\prod_{l = 0}^{l = n - 1} β_{n - l}} + \sum_{m = 0}^{m = n - 1} \frac{l n 2 - \frac{2}{2^{δ_{m}}}}{2^{\sum_{m = 0}^{m = n - 1} β_{m}}} > σ_{1} l n 2 \Rightarrow

and we see in case

δ_{m} > 2

With the growth of

δ_{m} > 2

, we see an exponential growth of the denominator, so the influence of

δ_{m} > 2

terms with large values does not have a significant effect on the estimate of the left side of the inequality. The reason is the accumulation of the

β_{m}

value corresponding to

δ_{m} = 1

. Therefore, the effect, taking into account

δ_{m} > 2

with large values, is insignificant for estimating the left part of the inequality. ⇒ statement of Theorem □

Theorem 4. Let

a_{n} = \sum_{i = 0}^{n} γ_{i} 2^{i}, n > 1000, γ_{i} \in {0, 1}

then

a_{4 n} < a_{n}

Proof. Let’s introduce operators defined formulas

P f = f / 2, T f = 3 f + 1, Z f = 3 f

T_{i} \in {P, T}, R_{i} \in {Z, P},

Let’s consider all possible scenarios of the behavior of the Syracuse sequence , the same possible scenarios can be written in the following form

\begin{matrix} a_{n + n} = T_{1} T_{2} \dots . T_{n} a_{n} \end{matrix}

We need to calculate an estimate for every 2n-th member of the Collatz sequence based on the number of applied operators P ,T,Z over n step.

\begin{matrix} a_{n + n} = T_{n} T_{n - 1} \dots . T_{1} a_{n} \end{matrix}

Let

a_{n}

have m ones in binary representation, then count the number of applications of the Z operator in the following formula.

m = \sum_{R_{i} = Z, i \leq n} 1

and count the number of applications of the P operator in the following formula.

\sum_{R_{i} = P, i \leq n} 1 = m + n - m

Because, each application Z is accompanied by an operator P and the number of applications of the operator P according to the zeros

a_{n}

has a corresponding n-m By rules of Collatz we have after n steps

a_{n + n} = 3^{m} / 2^{n} a_{n} + T_{n} T_{n - 1} \dots . T_{1} 1 = 3^{m} / 2^{n} a_{n} + B_{n}

B_{n} \leq \sum_{j = 1, m} 3^{j} / 2^{j} < 23^{m} / 2^{m} \leq 3 {(3 / 2)}^{m} 2^{- n} a_{n}

B_{n} \leq 3 {(3 / 4)}^{m} 2^{m - n} a_{n}

According to the last formula, we see that the growth of each member of the sequence depends on the number of units in the binary representation. Next, we will show that a large number of ones at the * step leads to an increase in the number of zeros in the binary representation according to the previous theorems. Where will the decrease in the following terms of the sequence follow

\begin{matrix} a_{2 n} = 3^{m} (a_{n} * 2^{- n} + B_{n}) = (a_{n} * 2^{- n} + B_{n}) 3^{m} \\ a_{2 n} = \sum_{i = 0}^{[α_{1}]} γ_{i} 2^{i}, γ_{i} \in {0, 1}, α_{1} = m * ln 3 / ln 2 + ln (2^{- n} a_{n}) \end{matrix}

Repeating the reasoning of Theorem 3, let’s consider the equation

2^{x} = a_{2 n} = 3^{m} a_{n} \cdot 2^{- n} + B_{n} = 3^{m} + 3^{m} (a_{n} - 2^{n}) \cdot 2^{- n} + B_{n}

x ln 2 = m ln (3) + ln (1 + (a_{n} - 2^{n}) \cdot 2^{- n} + B_{n} \cdot 3^{- m})

We come to the following

x ln 2 = m ln (3) + ln (1 + θ) + O (\frac{1}{2^{n} ln 2})

{x} = \{\frac{m ln (3)}{ln (2)} + \frac{ln (1 + θ)}{ln 2} + O (\frac{1}{2^{n} ln 2})\}

From the last equation, in order to apply the results of Theorem 3, we need

{x} > ln 2

. To execute the last inequality, consider the following cases

$γ_{n - 1} = 0, γ_{n - 2} = 0$
$γ_{n - 1} = 1, γ_{n - 2} = 1$
$γ_{n - 1} = 0, γ_{n - 2} = 1$
$γ_{n - 1} = 1, γ_{n - 2} = 0$

In addition, we accept the possibility of changing m

\Rightarrow m \pm 1

. The latter is possible by changing the number of applications of the Collatz rules or, in other words, by decreasing or increasing the elements of the sequence by one. As a result, we have

{x} = \{\frac{(m \pm 1) ln (3)}{ln (2)} + \frac{ln (1 + θ)}{ln 2} + O (\frac{1}{2^{n} ln 2})\}

Considering the following cases for

m = m, m = m - 1, m = m + 1

depending on the behavior of

γ_{n - 1}, γ_{n - 2}

we have the following three variants for the fractional part of x

{x} = \{\frac{(m \pm 1) ln (3)}{ln (2)} + \frac{ln (1 + θ)}{ln 2} + O (\frac{1}{2^{n} ln 2})\}

{x} = \{\frac{m ln (3)}{ln (2)} + \frac{ln (1 + θ)}{ln 2} + O (\frac{1}{2^{n} ln 2})\}

Thus, depending on the behavior of

γ_{n - 1}, γ_{n - 2}

we can always choose an option where the fractional part of x will satisfy the conditions of Theorem 3.

Denoting

\begin{matrix} m^{*} is the number of non - zero γ_{i} \\ l^{*} is the number of zero γ_{i} \end{matrix}

by theorem 3 we will have

\begin{matrix} m^{*} \leq [α_{1}] / 2 + 5 = [m ln 3 / ln 2] / 2 + 5 \\ l^{*} \geq [α_{1} / 2 - 5 = [m ln 3 / ln 2] / 2 - 5 \end{matrix}

After

3 n

steps of applying Collatz rules, we have

a_{4 n} = \frac{3^{m^{*}}}{2^{2 n}} a_{2 n} + T_{3 n} T_{3 n - 1} \dots . T_{1} 1 = \frac{3^{m^{*}}}{2^{2 n}} a_{2 n} + B_{3 n}

a_{4 n} = \frac{3^{m^{*}}}{2^{2 n}} a_{2 n} + T_{3 n} T_{3 n - 1} \dots . T_{1} 1 = \frac{3^{m^{*}}}{2^{2 n}} (\frac{3^{m}}{2^{n}} a_{n} + B_{n}) + B_{3 n}

a_{4 n} = 3^{m^{*} + m} 2^{- 3 n} a_{n} + 3^{m^{*}} 2^{- 2 n} B_{n} + B_{3 n}

B_{3 n} = 3^{m^{*}} 2^{- 2 n} B_{n} = 3^{m^{*} + m} 2^{- 2 n - n} a_{n}

a_{4 n} \leq q_{1} * a_{n}

by definitions

m^{*}, l^{*}, B_{n}

we get

q_{1} < 1

Using

n > 1000 \Rightarrow q_{1} < 1 \Rightarrow a_{4 n} < a_{n}

. □

Theorem 5. Let

a_{n} = \sum_{i = 0}^{n} γ_{i} 2^{i}, n > 1000, γ_{i} \in {0, 1}

then for

a_{n}

Collatz conjecture is true Proof. Proof follows from theorem 1-4

6. Conclusions

Our assertion proves that after

3 n

steps, a sequence with an initial binary length of n arrives at a number strictly smaller than the initial one, from which the solution to the Collatz conjecture follows. This is because by applying this process n times, we are guaranteed to arrive at 1.

References

O’Connor, J.J.; Robertson, E.F. "Lothar Collatz". St Andrews University School of Mathematics and Statistics, Scotland. 2006. [Google Scholar]
Tao, Terence. "Almost all orbits of the Collatz map attain almost bounded values". Forum of Mathematics, Pi. 10: e12. arXiv:1909.03562. ISSN 2050-5086. 2022. [Google Scholar] [CrossRef]

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